Answer:
The answer to your question is x ≠ -10 x ≠ 2 x ≠ -2
Solution [tex]\frac{(x- 8)(x + 2)}{(x + 8)(x - 2)}[/tex]
Step-by-step explanation:
Process
1.- Factor all the polynomials
[tex]\frac{x^{2} -3x - 40}{x^{2} + 8x - 20} / \frac{x^{2}+ 13x + 40}{x^{2}+ 12x + 20}[/tex] = [tex]\frac{(x-8)(x + 5)}{(x + 10)(x- 2)} / \frac{(x+ 8)(x + 5)}{(x + 10)(x + 2)}[/tex]
2.- Invert the second term
= [tex]\frac{(x- 8)(x + 5)}{(x +10)(x - 2)} \frac{(x+ 10)(x + 2)}{(x + 8)(x +5)}[/tex]
3.- Simplify
= [tex]\frac{(x- 8)(x + 2)}{(x + 8)(x - 2)}[/tex]
4.- x must be different to
x ≠ -10 x ≠ 2 x ≠ -2
This values are from the factor expressions and then equal to zero.
